Chapter 3: Moving From Individual Vectors to Spaces of Vectors
Earlier linear algebra focused heavily on individual vectors, linear combinations, matrices, elimination and systems such as \(Ax=b\). Chapter 3 begins asking a broader question:
What happens when we study an entire collection of vectors together?
The central idea is that a vector space must be stable under the linear-combination operations we already know.
What Does a Real Component Mean?
Consider
\[v=\begin{bmatrix}3\\5\end{bmatrix}.\]
The numbers \(3\) and \(5\) are the components, entries or coordinates of the vector.
A real number is any number belonging to the ordinary real number line, including integers, fractions, decimals and irrational numbers such as \(\sqrt2\) and \(\pi\).
The set of real numbers is denoted by
\[\mathbb R.\]
Therefore
\[\mathbb R^2\]
means all vectors containing two real components:
\[\boxed{\mathbb R^2=\left\{\begin{bmatrix}x\\y\end{bmatrix}:x,y\in\mathbb R\right\}}.\]
The superscript 2 does not mean that the numbers are squared. It means that each vector has two real components.
Similarly,
\[\mathbb R^3=\left\{\begin{bmatrix}x\\y\\z\end{bmatrix}:x,y,z\in\mathbb R\right\}.\]
Geometrically, \(\mathbb R^2\) is the entire two-dimensional plane and \(\mathbb R^3\) is ordinary three-dimensional space.
Connection to Quantum Computing
This distinction will eventually matter in quantum computing because quantum-state vectors can contain complex components. Instead of \(\mathbb R^n\), quantum mechanics commonly uses vector spaces over \(\mathbb C\).
For example,
\[\begin{bmatrix}\frac{1}{\sqrt2}\\\frac{i}{\sqrt2}\end{bmatrix}\]
belongs to \(\mathbb C^2\), not \(\mathbb R^2\), because \(i\) is not a real number.
From a Vector to a Whole Collection
Take the vector
\[v=\begin{bmatrix}1\\2\end{bmatrix}.\]
Consider every possible scalar multiple:
\[cv.\]
Examples include
\[-2v=\begin{bmatrix}-2\\-4\end{bmatrix},\qquad 0v=\begin{bmatrix}0\\0\end{bmatrix},\qquad 3v=\begin{bmatrix}3\\6\end{bmatrix}.\]
All of these vectors satisfy
\[y=2x.\]
Geometrically they form an entire line through the origin.
The Central Vector-Space Requirement
If \(v\) and \(w\) belong to a vector space \(S\), then every linear combination must also belong to \(S\):
\[\boxed{v,w\in S\quad\Longrightarrow\quad cv+dw\in S}.\]
The numbers \(c\) and \(d\) can be any real numbers.
The intuitive meaning is:
Take vectors from the space, scale them and add them in any way allowed by linear combinations. The result must never escape the space.
This property is often described by saying the space is closed under linear combinations.
The Railway-Track Picture
A useful mental picture is to imagine the subspace as a railway track.
If vectors \(v\) and \(w\) are on the track, then
\[cv+dw\]
must also land somewhere on the same track, regardless of the allowed values of \(c\) and \(d\).
If even one allowed linear combination leaves the track, the collection is not a vector subspace.
Why y = 2x Works
Consider
\[S=\left\{\begin{bmatrix}x\\y\end{bmatrix}:y=2x\right\}.\]
Any vector in this collection can be written as
\[v=\begin{bmatrix}a\\2a\end{bmatrix}.\]
Another can be written as
\[w=\begin{bmatrix}b\\2b\end{bmatrix}.\]
Now form an arbitrary linear combination:
\[cv+dw=c\begin{bmatrix}a\\2a\end{bmatrix}+d\begin{bmatrix}b\\2b\end{bmatrix}.\]
This becomes
\[\begin{bmatrix}ca+db\\2ca+2db\end{bmatrix}=\begin{bmatrix}ca+db\\2(ca+db)\end{bmatrix}.\]
If
\[k=ca+db,\]
then the result is
\[\begin{bmatrix}k\\2k\end{bmatrix}.\]
It still satisfies \(y=2x\). Therefore every linear combination remains on the same railway track.
So
\[\boxed{y=2x\text{ defines a subspace of }\mathbb R^2}.\]
Why Must a Vector Space Contain Zero?
Suppose \(v\) belongs to a vector space. Scalar multiplication must allow us to choose the scalar zero:
\[0v=\begin{bmatrix}0\\0\end{bmatrix}.\]
Therefore every vector space must contain the zero vector:
\[\boxed{0\in S}.\]
This immediately explains why vector subspaces represented geometrically as lines or planes must pass through the origin.
The Important Question: What Is Wrong With y = 2x + 1?
At first it seems that \(y=2x+1\) should work too. Its vectors can be written as
\[v=\begin{bmatrix}a\\2a+1\end{bmatrix},\qquad w=\begin{bmatrix}b\\2b+1\end{bmatrix}.\]
The natural question was: if both vectors are on the same railway track, shouldn’t their linear combinations remain there?
Let’s test it rather than relying on appearance.
Using the same scalar \(c\) for both vectors:
\[cv+cw.\]
This gives
\[c\begin{bmatrix}a\\2a+1\end{bmatrix}+c\begin{bmatrix}b\\2b+1\end{bmatrix}.\]
The first component is
\[c(a+b).\]
The second is
\[c(2a+1)+c(2b+1)=2c(a+b)+2c.\]
If the first component is called \(x\), then the resulting vector satisfies
\[y=2x+2c,\]
not necessarily
\[y=2x+1.\]
Therefore the linear combination can leave the original line.
A Very Simple Counterexample
Choose two vectors on \(y=2x+1\):
\[v=\begin{bmatrix}0\\1\end{bmatrix},\qquad w=\begin{bmatrix}1\\3\end{bmatrix}.\]
Both satisfy the equation.
But their sum is
\[v+w=\begin{bmatrix}1\\4\end{bmatrix}.\]
For \(x=1\), the original line requires
\[y=2(1)+1=3,\]
but the resulting vector has \(y=4\).
Therefore
\[\boxed{y=2x+1\text{ is not closed under linear combinations}.}\]
It is a line in \(\mathbb R^2\), but it is not a vector subspace.
Why the +1 Causes Trouble
For \(y=2x\), adding vectors preserves the relationship:
\[2a+2b=2(a+b).\]
But for \(y=2x+1\), adding two vectors adds the constant terms too:
\[(2a+1)+(2b+1)=2(a+b)+2.\]
The required \(+1\) has become \(+2\). The fixed displacement from the origin is not preserved by ordinary linear combinations.
Whole Space Versus Subspace
An important confusion arose with the vector
\[\begin{bmatrix}1\\100\end{bmatrix}.\]
It clearly does not satisfy \(y=2x\), because
\[100\ne2(1).\]
Therefore it does not belong to the particular subspace \(S\).
But it still belongs to \(\mathbb R^2\), because \(\mathbb R^2\) contains every vector with two real components.
Thus
\[\boxed{S=\{(x,y):y=2x\}\subset\mathbb R^2}.\]
The entire plane \(\mathbb R^2\) is the larger vector space. The line \(S\) is a smaller vector space contained inside it.
Does a Subspace Mean y Must Depend on x?
For the particular subspace
\[y=2x,\]
yes: once \(x\) is selected, \(y\) is no longer independently selectable.
For example,
\[x=3\quad\Longrightarrow\quad y=6.\]
Every vector in this subspace can therefore be written as
\[\begin{bmatrix}x\\2x\end{bmatrix}=x\begin{bmatrix}1\\2\end{bmatrix}.\]
Although the vector contains two components, there is only one freely chosen number. Choosing \(x\) automatically determines \(y\).
This is why the subspace geometrically forms a line and has only one independent direction.
The More General Insight
It would be too restrictive to say that every subspace requires \(y\) specifically to depend on \(x\).
A better statement is:
\[\boxed{\text{A subspace may impose constraints that reduce the number of independent choices.}}\]
For example, in \(\mathbb R^3\), the constraint
\[x+y+z=0\]
can be rewritten as
\[z=-x-y.\]
Here \(x\) and \(y\) could be chosen freely, while \(z\) would then be determined.
This observation points toward an important idea that will appear throughout Chapter 3: dimension measures the independent freedom available inside a vector space or subspace.
Current Mental Model
At this stage, the important picture is:
\[\boxed{\mathbb R^2=\text{all two-component real vectors}}\]
while a subspace such as
\[\boxed{S=\{(x,y):y=2x\}}\]
contains only vectors satisfying its constraint.
The decisive test is not merely whether the set looks like a line or plane. The decisive question is:
If I take vectors from this set and form arbitrary linear combinations, can I ever escape?
If the answer is no, the set is closed under linear combinations and can be a vector subspace.
Note metadata
- Note type: learning-note
- Subject: learning-notes
- Source: Introduction to Linear Algebra, Fifth Edition